Question 251139
It might be easier to understand the solution if we break it into two parts:<ol><li>Figure out what {{{3^x}}} is</li><li>Use part 1 to find what x is</li></ol>
To solve for {{{3^x}}} I am going to use a temporary variable:
Let {{{q = 3^x}}}
Now our equation is:
{{{q/(q+3) = 1/3}}}
We can eliminate the fractions by multiplying by the lowest common denominator (which is 3*(q+3)):
{{{3*(q+3)(q/(q+3)) = 3*(q+3)(1/3)}}}
On the left the (q+3)'s cancel and on the right the 3's cancel leaving:
{{{3*q = (q+3)*1}}}
or
{{{3q = q+ 3}}}
Subtract q from each side:
{{{2q = 3}}}
Divide both sides by 2:
{{{q = 3/2}}}
Replace q with {{{3^x}}}
{{{3^x = 3/2}}}
(We could have solved for {{{3^x}}} without the temporary variable but most students find it a little harder to follow.) Now that we know what {{{3^x}}} is we can solve for x. Since 3/2 is not an obvious power of 3 and since x is in an exponent we will use logarithms. If you do not need any decimal approximation for the answer then base 3 logarithms would be best. (I'll show you later.) In case you do need a decimal approximation use a logarithm your calculator can find, like base 10:
{{{log((3^x)) = log((3/2))}}}
Now we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front. This is one way to get variables out of exponents and the reason we use logarithms to solve these equations:
{{{x*log((3)) = log((3/2))}}}
Divide both sides by log(3):
{{{x = log((3/2))/log((3))}}}
This is an exact answer for x. If you need a decimal approximation for the answer then find the two logarithms and then divide them.<br>
If you use base 3 logarithms we would end up with:
{{{x = log(3, (3/2))/log(3, (3))}}}
Since {{{log(3, (3)) = 1}}} by definition we get:
{{{x = log(3, (3/2))/1}}}
which is
{{{x = log(3, (3/2))}}}
And this will simplify even further if we use another property of logarithms, {{{log(a, (p/q)) = log(a, (p)) - log(a, (q))}}}:
{{{x = log(3, (3)) - log(3, (2))}}}
{{{x = 1 - log(3, (2))}}}
which is a much simpler exact answer. But getting a decimal approximation for this would be a bit of a pain since there are no calculators I know of which can find base 3 logarithms. (You could use the base conversion formula, however.)